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If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3)

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If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) [#permalink]

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If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

A. 25
B. 27
C. 45
D. 75
E. 125
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Nov 2013, 13:56, edited 2 times in total.
Renamed the topic and edited the question.

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Re: Factor/exponent question [#permalink]

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john4 wrote:
If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

A.) 25
B.) 27
C.) 45
D.) 75
E.) 125


I don't understand how to tackle this question. Thanks!


Hi John,

Solution :
\((2^5) * (6^2) * (7^3) * n\) is the given number.

If both 5^2 & 3^3 are factors, then they must be present in the number.

Leaving rest of the prime factors and splitting 6^2 into 3^2 * 2^3.

The number is lacking 5^2 & a 3, so that 5^2 and 3^3 is a factor.

Hence the smallest number is 5^2 * 3 = 75

Hope it is clear

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Re: Factor/exponent question [#permalink]

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New post 17 Nov 2013, 13:55
Hello qoofi,

Please explain me more clearer.
Please check I dont get the answer of this question,


If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) :

It means that n*2^5*6^2*7^3 his number is dicvisible by these 2 factors.
If it is divisible so we can write the equation as,

n*5^2*6^2*7^3/5^2*3^3
= n*5^2*(2^2*3^2) *7^3/5^2*3^3=so after this how to get the answer I dont get it.
Please explain me.

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Re: Factor/exponent question [#permalink]

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New post 26 Nov 2013, 09:19
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sumit88 wrote:
Hello qoofi,

Please explain me more clearer.
Please check I dont get the answer of this question,


If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) :

It means that n*2^5*6^2*7^3 his number is dicvisible by these 2 factors.
If it is divisible so we can write the equation as,

n*5^2*6^2*7^3/5^2*3^3
= n*5^2*(2^2*3^2) *7^3/5^2*3^3=so after this how to get the answer I dont get it.
Please explain me.


First, if a number, lets say X, is a factor of another number, lets say Y . Then Y is divisible by X

In the question, both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), so n x (2^5) x (6^2) x (7^3) must be divisible by 5^2 and 3^3

we can rewrite n x (2^5) x (6^2) x (7^3) as n x (2^5) x (2^2) x (3^2) x (7^3)

so to make (2^5) x (2^2) x (3^2) x (7^3) divisible by 5^2 and 3^3, we need 5^2 and 3^1 ( if Y is divisible by X, then all prime factors of X must also be prime factors of Y)

so n= 5^2 and 3^1 which equals 75.

hope you find it helpful.
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Re: If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) [#permalink]

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New post 26 Feb 2014, 01:46
Just compare 5^2 and 3^3 with >>>>>>>> (2^5) x (6^2) x (7^3)

5^2 = 25 is not present; also 3^2 is present instead of 3^3, so one additional 3 is required as well

So, n= 25x3 = 75 = Answer = D
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Re: If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) [#permalink]

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Re: If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) [#permalink]

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New post 10 Sep 2017, 07:13
john4 wrote:
If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

A. 25
B. 27
C. 45
D. 75
E. 125

Given \(n(2^5)(6^2)(7^3)\) is divisible by both \(5^2\) and \(3^3\).

Therefore \(n(2^5)(6^2)(7^3)\) will be divisible by LCM of \(5^2\) and \(3^3\).

LCM of \(5^2\) and \(3^3 = (5^2)(3^3)\)

By Simplifying the expression we get;

\(\frac{n(2^5)(6^2)(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^5)(2*3)^2(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^5)(2^2)(3^2)(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^7)(3^2)(7^3)}{(5^2)(3^3)}\)

\(\frac{n(2^7)(7^3)}{(5^2)(3)}\)

\(2\) and \(7\) are not divisible by \(5\) or \(3\). Hence \(n\) should be a multiple divisible by \((5^2*3)\)

Therefor \(n(2^7)(7^3)\) to be divisible by \((5^2*3)\), smallest possible of \(n\) Should be \(= 5^2*3 = 25*3 = 75\)

Answer (D)...

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Re: If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3)   [#permalink] 10 Sep 2017, 07:13
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